Question

integrate the following

Answer 1

\[\int\limits_{0}^{1} \frac{ x^a-x^b }{ \log_{e}x } dx\]

Answer 2

the answer is \[\log_{e} \frac{ (a+1) }{(b+1) }\]

Answer 3

@satellite73 @amistre64

Answer 4

@ganeshie8

Answer 5

@Hero

Answer 6

\[\Large \int_{0}^{1} \frac{ x^a-x^b }{ \ln x } dx\]
is there any certain techniques? or is this spose to be done with basic calc2 stuff?

Answer 7

anyhow......???

Answer 8

thanks for responding to the call

Answer 9

i go no idea how "proper" this is but:\[y=\frac{ x^a-x^b }{ \ln x } \]
\[y\ln(x)=x^a-x^b\]
\[\large \int x^{y}dx=\int e^{x^a-x^b}dx\]
\[\large \int x^{y}dx=\frac{1}{a~x^{a-1}-b~x^{b-1}}e^{x^a-x^b}\]
wondering if we can just do x^(y+1)/(y+1) ... but thats prolly improper

Answer 10

@mathstudent55 can you come and help me

Answer 11

i think the integration you did is quite wrong......

Answer 12

can we back engineer it?

\[ln \frac{ (a+1) }{(b+1) }\to ln(f(b))-ln(f(a))\]
\[ln(f(x))\to \frac{f'(x)}{f(x)}=\frac{x^a-x^b}{ln(x)}\]

Answer 13

well,that is; f(1) = a+1, f(0) = b+1

Answer 14

\[\frac 1x = x^a+x^b\]??? just thinking

Answer 15

Oh you've got me thinking amistre!
Could we do something like this maybe?
Introduce an x in the top and bottom.

\[\large \int\limits\frac{x^{a+1}+x^{b+1}}{x \ln x}dx\]

Now the bottom of the fraction works out a lot better.
Maybe allowing us to do by-parts or something maybe...

Answer 16

thats what i was thinking :) and its a subtraction up to of course

Answer 17

Oh woops :3

Answer 18

we can split the top if needed to make 2 integrals ....

Answer 19

we still have some issue with the limits of integration, so theres goung to be some improper integral limit to be had im sure

Answer 20

\[\large u= \ln x \qquad\qquad \qquad du=\frac{1}{x}dx\]\[\large x=e^u\]

\[\large \int\limits_{x=0}^1 \frac{x^{a+1}-x^{b+1}}{x \ln x}dx \qquad\to\qquad \int\limits_{x=0}^1 \frac{e^{u(a+1)}-e^{u(b+1)}}{u}du\]
We could integrate by parts from u here?
Oh the limits :OOOO interesting...

Answer 21

that does look to be headed in the right direction now :)

Answer 22

when x=0, u = -inf; when x=1, u = 0 and then we dont have to come back to xs :)

Answer 23

no it comes out to be infinity.......!

Answer 24

i am pretty sure there must be some computational gimmick, because you are not going to find a closed form anti derivative for this

Answer 25

what context is this question in?

Answer 26

ok i worked out a method like this but don't know its validity............\[\int\limits_{0}^{1}\frac{ x^a-x^b }{\log_{e}x }=\int\limits_{0}^{1} { \int\limits_{a}^{b}x^y dy }dx\]

Answer 27

now we can change the order of integration and hence obtain an answer;...

Answer 28

wow that looks good!

Answer 29

might need to switch \(a\) and \(b\) though

Answer 30

is that correct.....

Answer 31

\[\int_b^ax^ydy=\frac{x^a-x^b}{\log(x)}\] is right for sure

Answer 32

good work!

Answer 33

swapping the integration dont seem to produce the end result tho; but that was pretty neat ... kinda what my first idea morphed into, but different :)